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STRUCTURE OF SODIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ”(2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. There are twenty recognized isotopes of sodium, ranging from Na-18 to Na-37. Na-23 is the only stable (and the only primordial) isotope. As such, it is considered a monoisotopic element and it has a standard atomic mass: 22.98976928(2) u. Sodium has two radioactive cosmogenic isotopes (Na-22, half-life = 2.605 years; and Na-24, half-life ≈ 15 hours). With the exception of those two, all other isotopes have half-lives under a minute, most under a second. The least stable is Na-18, with a half-life of 1.3(4)×10−21 seconds. Acute neutron radiation exposure (e.g., from a nuclear criticality accident) converts some of the stable Na-23 in human blood plasma to Na-24. By measuring the concentration of this isotope, the neutron radiation dosage to the victim can be computed.Sodium-22 is a positron-emitting isotope with a remarkably long half-life. It is used to create test-objects and point-sources for positron emission tomography. WHY Na-23 IS A STABLE NUCLIDE After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per neutron or per proton. In the case of Na-23 (See my STRUCTURE OF Na23 AND Ne23 ) using the diagram of Na-23 we see that the p11 makes two very strong vertical bonds with two neutrons existing outside the parallelepiped (core), while in the unstable Na-22 the p11 and p10 make a vertical rectangle outside the parallelepiped with two squares in which the pp systems of parallel spin give very strong magnetic repulsions. On the other hand in the isotopes with extra neutrons the single pn bonds of horizontal weak binding energies lead to the beta decay. ' ' ' DIAGRAM OF Na-23 WITH S = +3/2' Here the n11, p11 and n12 make vertical strong bonds outside the parallelepiped (core) formed by 10 protons and 10 neutrons. It this structure the first horizontal plane has four nucleons of positive spins (+HP1) . The total spin S =+3/2 is due to the 13 nucleons of positive spins and to the 10 nucleons of negative spins. ' p10(+1/2.n10(+1/2) ' ' n9(+1/2).p9(+1/2) +HP5 ' ' n8(-1/2).. p8(-1/2)…..n12(-1/2) ' ' p7(-1/2).n7(-1/2) - HP4 ' ' p6(+1/2).n6(+1/2)…..p11(+1/2) ' ' n5(+1/2).p5(+1/2) + HP3 ' ' n4(-1/2).p4(-1/2)……n11(-1/2) ' ' p3 (-1/2).n3 (- 1/2) - HP2 ' ' p2(+1/2).n2(+1/2) ' ' n1(+1/2). p1(+1/2) + HP1 ' ' ' STRUCTURE OF Na-22, Na-21, Na-20, Na-19, Na-18 ''' After a careful analysis of these nuclides I found that the structure of all the above nuclides have a structure based on the structure of Na-22 with S = +3. In the following diagram of Na-22 you see that it consists of a parallelepiped with 18 nucleons giving S = +3 in which an additional rectangle of p10n10p11n11 with unstable bonds has a total spin S = 0. In other words the nucleons of the first horizontal plane (+HP1) and the second horizontal plane (-HP2) give S= 0, while the nucleons of the third horizontal plane (+HP3) give S =+3. '''DIAGRAM OF Na-22 WITH S = +3 ' n6(+1/2).p6(+1/2)n9(+1/2) ' ' p5(+1/2).n5(+1/2)p9(+1/2) + HP3 ' ' p4(-1/2).n4(-1/2)p8(n-1/2 )n11(-1/2) ' ' n3 (-1/2).p3 (- 1/2)n8(-1/2).p11(-1/2) - HP2 ' ' n2(+1/2).p2(+1/2)n7(+1/2).p10(+1/2) ' ' p1(+1/2). n1(+1/2).p7(+1/2).n10(+1/2) + HP1 ' In the absence of a neutron like the n9(+1/2) we get the structure of Na-21 with S = +3/2 because the p9(+1/2 ) as p9(-1/2) goes to the n3(-1/2) in order to make a weak horizontal bond with n3. Thus the +HP3 has 4 nucleons of positive spins , while the -HP2 has 9 nucleons of negative spins. Since the +HP1 has 8 nucleons of positive spins we get S = 4(+1/2) + 9(-1/2) + 8(+1/2) = +3/2. In the absence of two neutrons like the n6(+1/2) and n10(+1/2) which are at the corners of the parallelepipeds we get the structure of Na-20 with S =+2 because S = +3 -(+1/2) - (+1/2) = +2. Now taking into account the structure of Na -20 with S =+2 we get the structure of N-19 with S = +5/2 when the third absent neutron is the n11(-1/2) of the corner of the parallelepiped. Thus S = +2 - (-1/2) = +4/2 +1/2 = +5/2 Finally in the absence of 4 neutrons the structure of N-18 with S = -1 is based on the same structure of Na-22 when all the nucleons change the spins. For example in this case we have -HP1, +HP2 and -HP3 with a total S = -3. So in the absence of four neutrons of negative spins we get S = -3 -4(-1/2) = - 6/2 + 4/2 = -2/2 = -1. STRUCTURE OF Na-24, Na-25, Na-26, Na-27, Na-28, Na-29, Na-30, Na-31, Na-32, Na-33, Na-34, Na-35, Na-36, Na-37. Also in these cases the structure is based on the structure of Na-22 with S = +3. For example adding two extra neutrons of positive spins which make weak horizontal bonds with protons we get the structure o Na-24 with S = +4. That is S= +3 + 2(+1/2) = +4 Similarly in the structure of Na-25 with S = +5/2 we see that the 2 of the 3 extra neutrons have negative spin, while the third extra neutron has a positive spin. That is S = +3 + 2(-1/2) + 1/2 = +5/2 . However in the structure of Na-26 with S = +3 the 4 extra neutrons have opposite spins, while in Na-27 with S = +5/2 the 3 of the 5 extra neutrons have negative spins and the 2 have positive spins. That is S = +3 + 3(-1/2) + 2(+1/2) = +5/2 . In the structure of Na-28 with S = +1 we see that 5 extra neutrons have negative spins and one extra neutron has a positive spin. That is S = +3 + 5(-1/2) +1/2 = +1. In the structure of Na -29 with S = +3/2 having 7 extra neutrons we see that it has 5 extra neutrons of negative spins and 2 extra neutrons of positive spins. That is S = +3 +5(-1/2) + 2(+1/2) = +3/2. In the structure of Na-30 with S = +2 having 8 extra neutrons we see that it has 6 extra neutrons of negative spins and 2 extra neutrons of positive spins. That is S = +3 +6(-1/2) + 2(+1/2) = +2 Following the same method one can find the number of extra neutrons with positive or negative spins for the nuclides like the Na-31 with S = +3/2, the Na-33 with S = +3/2, the Na-34 with S = +1 the Na-35 with S= +3/2 and the N-37 with S = +3/2. However in the case of Na-32 with S = -3 we see that it is based on the same structure of Na-22 when the nucleons change their spins. For example in this case we have -HP1, +HP2 and -Hp3 with a total spin S = -3. So in the structure of Na-32 with S = -3 we have 10 extra neutrons of opposite spins giving a total S = 0.That is S = -3 +5(+1/2) +5(-1/2) = -3. Category:Fundamental physics concepts